Can a concept satisfaction of an expression, contains both type and the reference?

user_1438465 0
Viewed 129

Is there a way to make the following code not so bloated?

I mean join both type and a reference somehow (|| does not work).

template<typename T>
concept IntegralVector = std::integral<typename T::value_type> &&
requires(T t)
{
    { t.size() } -> std::convertible_to<std::size_t>;
}
&& (requires(T t)
{
    { t[0] } -> std::same_as<typename T::value_type&>;
} || requires(T t)
{
    { t[0] } -> std::same_as<typename T::value_type>;
});

A working trick can be:

{ 0 + t[0] } -> std::integral;

But I want to stick with typename T::value_type

c++20 c++
2 Answers

You probably want something like this:

template <typename T, typename U>
concept decays_to = std::same_as<std::decay_t<T>, U>;

To use as:

template<typename T>
concept IntegralVector =
    std::integral<typename T::value_type>
    && requires (T t) {
        { t.size() } -> std::convertible_to<std::size_t>;
        { t[0] } -> decays_to<typename T::value_type>;
    };

This also catches value_type const& as an option, which I'm not sure was intentionally omitted.

user_20690640

Another alternative is to use C++23 auto(x), which is the decay-copy in the language:

#include<concepts>

template<typename T>
concept IntegralVector = std::integral<typename T::value_type> &&
requires(T t) {
  { t.size() } -> std::convertible_to<std::size_t>;
  { auto(t[0]) } -> std::same_as<typename T::value_type>;
};

Demo.

user_116387180
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